Log2r bits per step. This is what is meant, essentially, by different
Transducers having different “capacities” for transmission.
Conversely, as S’s variety mounts step by step we can see that
The amount of variety that a transducer (such as R) can transmit
Is proportional to the product of its capacity, in bits, and the
Number of steps taken. From this comes the important corollary,
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Which will be used repeatedly later: given long enough, any trans-
Ducer can transmit any amount of variety.
An important aspect of this theorem is its extreme generality.
What sort of a machine it is that is acting as intermediate trans-
Ducer, as channel, is quite irrelevant: it may be a tapping-key that
Has only the two states “open” and “closed”, or an electric poten-
Tial that can take many values, or a whole neural ganglion, or a
Newspaper— all are ruled by the theorem. With its aid, quantita-
Tive accuracy can be given to the intuitive feeling that some
Restriction in the rate of communication is implied if the commu-
Nication has to take place through a small intermediate transducer,
Such as when the information from retina to visual cortex has to
Pass through the lateral geniculate body, or when information
About the movements of a predator have to be passed to the herd
Through a solitary scout.
Ex. 1: An absolute system, of three parts, Q, R and S, has states (q,r,s) and trans-
Formation
q:1 2 3 4 5 6 7 8 9q': ↓ 4 6 6 5 6 5 8 8 8
0,q + r even,r' = 1, if ,, is odd.,,,,
s' = 2s – r.
Q thus dominates R, and R dominates S. What is R’s capacity as a channel ?
Ex. 2: (Continued.) Nine replicates were started at the initial states (1,0,0),
So that only Q had any initial variety. (i) How did the
Variety of the Q’s change over the first five steps? (ii) How did that of the
R’s ? (iii) That of the S’s ?
Ex. 3: (Continued.) Had the answer to Ex. 2(iii) been given as “S:1,1,4,5,5”, why
Would it have been obviously wrong, without calculation of the actual trajec-
Tories ?
The exercise just given will have shown that when Q, R and
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S form a chain, S can gain in variety step by step from R even
Though R can gain no more variety after the first step (S.8/12). The
Reason is that the output of R, taken step by step as a sequence,
Forms a vector (S.9/9), and the variety in a vector can exceed that
In a component. And if the number of components in a vector can
Be increased without limit then the variety in the vector can also
Be increased without limit, even though that in each component
Remains bounded. Thus a sequence of ten coin-spins can have
Variety up to 1024 values, though each component is restricted to
Two. Similarly R’s values, though restricted in the exercises to
Two, can provide a sequence that has variety of more than two. As
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A N I N T R O D UC T I O N T O C Y B E R NE T I C S
T RA N SMISSI O N O F VA R IE TY
The process of transmission goes on, S is affected (and its variety
Increased) by the whole sequence, by the whole vector, and thus a
Variety of much more than two can pass through R. A shrinkage
In the capacity of a channel can thus be compensated for (to keep
The total variety transmitted constant) by an increase in the length
Of the sequence— a fact already noticed in the previous section,
And one that will be used frequently later.
Ex. 1: An absolute system T dominates a chain of transducers A1, A2, A3, A4,…:
T → A 1 → A2 → A3 → A4 → …
A set of replicates commences with variety in T but with none in A1, nor in
A2, etc. Show that after k steps the varieties in A1, A2, . . ., Ak may be
non-zero but that those in Ak+1, Ak+2, . . . must still be zero (i.e. that T’s vari-
Ety “cannot have spread farther than Ak”.).
Ex. 2: Of 27 coins, identical in appearance, one is known to be counterfeit and to
Be light in weight. A balance is available and the counterfeit coin is to be
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Identified by a series of balancings, as few as possible. Without finding the
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